 ×     #### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```
find the no. of solution of -x^2+x-1 = x^2008 + |x|

```
11 years ago

```							hi rahul,
dis ques can be done by d approach of graph,,,
for LHS,
f = -x^2+x-1       ,f < 0 for all x <0,n f<0 for x>1, f >0 for 0<x>1........

since dy/dx < 0, n dis is a monotonus fxn so it decreases from +infinite to o at x=1,,,,,,,,,,
for rhs, f = x^2008 + |x|,   dis is a increasing fxn,,,,,n for all x f>0....
at x=1 , f=2,,at x =0,f=0,,,
so d two fxn can only intersect at one point in b/w 0<x>1...
so no. of soln = 1.
```
11 years ago
```							for x>0,
-x^2-1=x^2008
x^2008+x^2+1=0
So no positive roots (since all are positive terms)
for x<0,
-x^2+2x-1=x^2008
x^2008+(x-1)^2=0
so no negative roots (since all are positive terms)
So i think number of solutions is zero
```
10 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Algebra

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions